I'm a Science Olympiad coach trying to optimize the performance of our "Scrambler", a car which must be accelerated by only a falling mass.

Most competitors simply tie a weight to a string and route that string over a set of pulleys (using no mechanical advantage to convert the vertical falling acceleration horizontal. Even if the car had no mass and there were no loss of energy due to friction, that would limit the final departure velocity of the car to the final velocity of the falling weight:

v(final) = SQRT(2*g*h) = SQRT(2*9.8*1) = 4.427 m/s Intuitively, we believe we can get the car to depart faster if we store much of the KE (Kinetic Energy) of the falling mass into PE of the stretched rubber band, which would allow us to accelerate a car of much less mass at a rate considerable more than g (gravity).

Here's are target configuration, so far:

Mass to drop: 2Kg (max allowable)

Height of fall: 1 meter (max allowable) Mass of car we will accelerate:

0.5 Kg Max length of our launch device: 1 meter Useful details about our launch device:

1) Held at 1m high, our 2Kg mass is attached to a string that goes over a "top pulley'

2) Down on the floor, sits our car, which (for simplification) is 1 meter long.

3) Directly overhead and inline with the length of the car is a suspended 'slide rod' running the same 1 meter length of the car.

4) Hanging down from the rod is a sliding 'push block'. The bottom of this push block touches the 'back bumper' of our car, ready to push the car forward as it slides forward on the overhead rod. (NOTE: the actual length of the 'push' doesn't need to be a full 1 meter, and likely won't be)

5) Connection between falling mass and car: A string is attached to the 2Kg mass, immediate goes up and over a 'top pulley' then extends down to a 'bottom pulley' right at the front of the car and the front end of the slide rod. The string then continues horizontally to the other end of the slide rod over the back end of the car and is attached to a 'push block'

6) The 'push block' hangs from and slides along the slide rod and extends down just behind the 'back bumper' of the car. When the string gets pulled, the slide block is pulled forward, in turn, pushing the car forward as the 'launch'.

That was all simple, but needed as base information. Here's where it gets interesting!

1) The car is LOCKED into place and is released ONLY AFTER the falling mass arrives at the BOTTOM of its fall.

2) Very important: Our connection 'string' is not ALL string. inserted in between two end lengths is a strong rubber band.

3) Also very important: The string connection is 'slack' enough that the band doesn't not begin to stretch until the mass has fallen a significant % of the falling distance (we can adjust this). in this way we can allow the falling mass to accumulate KE (Kinetic Energy) before 'capturing' and storing that energy at PE in the rubber bank).

4) Through adjusting both the strength of the rubber bank and the % of the falling distance we use for KE-to-PE storage, we in tend to decelerate the falling weight back to near-zero by the time is reaches the floor. In this way, we will hopefully maximize the transfer of energy into the rubber band.

5) With the slight amount of inertia left in the falling/decelerated mass, this mass will fall onto a 'trap' that a) captures it, so it can't proceed back upward in response to the rubber band's 'pull', and b) releases the trigger holding car in place, thus allowing the stored energy to now pull on the slide block, pushing and accelerating the car.

HOW WE THING WE CAN OUT-PERFORM (OUT-ACCELERATE) OUR COMPETITION: By waiting to accumulate significant KE from the fall before converting it to PE in the band, we believe we can store a significant stretch force to accelerate the car. And, since our car is only 1/4 the mass of the falling weight, we should be able to use that force to accelerate the at a rate much higher than gravity's 9.8 m/s^2.

WHERE WE NEED YOUR HELP: Obviously, if we can get our heads around all the necessary formulae, we can calculate the resulting FINAL LAUNCH VELECITY OF THE CAR, and by experimenting with different condition variables, we can select the configuration for optimal performance.

So, getting to that calculation of Final Launch Velocity is our goal, and the target of our request for help.

Here are some of the thinking we've gone through, the formulae we've used, and some of the control limits we've settled on to keep our variables from getting out of hand:

1) We know we're starting with a PE = mgh = 2x9.8x1 = 19.6 Joules

2) We know the most straightforward way to increase our potential for faster acceleration is to lighten the car, but let's assume that 1/2 Kg is as good as we can get.

3) WE THINK WE'RE OUR THINKING IS SOUND HERE: Let's assume that we could precisely adjust the appropriate MOE (modulus of elasticity) and length of the rubber band so that we could always get the falling weight to decelerate back to zero just as it got to the floor, no matter when we allow the string to become taut and the band to begin to stretch. We're thinking that, ignoring friction and other inefficiencies, we'd always end up storing the same 19.6 Joules of PE in the rubber band. Two examples:

a) Weight falls .5 meter before band begins to stretch:

V(final) = SQRT(2 x g x distance) = 2 x 9.8 x 0.5 = 3.13 m/s

Accumulated KE = 1/2 x m x v^2 = 1/2 x 2 x 3.13m/s = 9.8 Joules

Remaining PE = mgh = 2 x 9.8 x 0.5 = 9.8 Joules.

Assuming band strength to decelerate the mass back to 0 when

reaching the floor, the energy stored in the band = 19.6 Joules

b) Weight falls .75 meter before band begins to stretch:

V(final) = SQRT(2 x g x distance) = 2 x 9.8 x 0.75 = 3.834 m/s

Accumulated KE = 1/2 x m x v^2 = 1/2 x 2 x 3.834m/s = 14.7 Joules

Remaining PE = mgh = 2 x 9.8 x 0.25 = 4.9 Joules.

Assuming band strength to decelerate the mass back to 0 when

reaching the floor, the energy stored in the band = 19.6 Joules

So, unless we've miscalculated, we believe there's no "best time" to start capturing energy into the band, assuming, again, that we capture all the energy by perfectly decelerating the falling mass back to zero at the bottom.

4) NOW, HERE'S A POSSIBLE AREA OF MISPERCEPTION:

While we've concluded that different 'stretch length / MOE' combinations will all result in the same 19.6 Joule energy transfer to our rubber band in our 'ideal' machine', it's also become obvious that, the longer we wait before beginning to stretch the rubber band, the greater will be the INITIAL force available to us to use to accelerate the car, and we're not sure if that will make any difference in how fast we can get our car moving.

HERE'S THE TWO SCENARIOS:

1) Example 1: the 2Kg mass falls 75cm (.75m) before the string goes taut and the 'well-adjusted' band slows the mass to a stop at zero height, having

stored 19.6 PE Joules. When this energy is used to accelerate the car, all

19.6 Joules will be expended within the 25 CENTIMETERS of 'un-stretching' it take to bring the band back to a 'limp' condition

2) Example 1: the 2Kg mass falls 50cm (.5m) before the string goes taut and the 'well-adjusted' band slows the mass to a stop at zero height, having

stored 19.6 PE Joules. When this energy is used to accelerate the car, all

19.6 Joules will be expended within the 50 CENTIMETERS of 'un-stretching' it take to bring the band back to a 'limp' condition.

To us, it just stands to reason that the magnitude of pulling force available from the shorter band will be much greater than the initial force available from the longer band. Since the shorter band will release/transfer all of it 19.6 Joules of energy in just 25cm while the longer band will take 50cm to do that,

That larger initial force pulling the 1/2 Kg car will, at first, impart a greater initial acceleration, but will dissipate more quickly.

THINKING...

Given that the slide rod is 1 meter long, we can keep pulling the car until for the full distance needed to "un-stretch" the band back to rest, regardless of whether we use 25cm stretch or the 50cm stretch. So, ultimately, ignoring inefficiencies (friction, etc.), the ultimate WORK available to be performed on the car is the same, derived from the 19.6 Joules of energy. But, somehow, we're not sure that necessarily means both stretch lengths will impart the same acceleration and highest launch velocity as one another. In the past, with similar but different car-competition events, we've observed that the more quickly we can impart acceleration force on our car, the greater the ultimate final launch velocity we can accomplish.

SO, OUR QUESTION IS THIS: Given the greater initial pulling force available from the shorter, stronger 25cm-stretch band and the resulting quicker initial acceleration, would we be able to reach a greater launch velocity that with the weaker 50cm-stretch band, or (ignoring inefficiencies) will the final launch velocities be the same?